1 EXAMPLES. 475 1. Required to find whether the decimal equal to be finite or infinite, and if infinite, how many places that repetend will consist of. First 25 Then, 475 2500 7)999999 19 =- (2) (2) (2) HR 2) 112 112 ; therefore, because the denominator 112 did not vanish in dividing by 2, the decimal is infinite, and, as six 9s were used, the circulate consists of 6 places, beginning at the fifth place, because four 2s were used in dividing. 2. Let be the fraction proposed. 3. Let be the fraction proposed. ADDITION OF CIRCULATING DECIMALS. RULE. 1. Make the repetends similar and conterminous, and find their sum as in common addition. 2. Divide this sum (of the repetends only) by so many nines as there are places in the repetend, and the remainder is the repetend of their sum; which must be set under the figures added, with cyphers on the left hand, when it has not so many places as the repetends. 3. Carry the quotient of this division to the next column, and proceed with the rest as infinite decimals. EXAMPLES. 1. Let 5·3+59-4356+3976+519+39+217·5 be added together. 1199.3851305 the sum. In this question, the sum of the repetends is 2851303, which divided by 999999, gives 2 to carry to the next column 5,3,0, &c. and the remainder is 851305. 2. Let 3275-319+36-45+123-19+5·3173+112-3513+11-131 +125+29.10053 be added together. 00042. Ans. 3593-00042. SUBTRACTION OF CIRCULATING DECIMALS. RULE. Make the repetends similar and conterminous, and subtract as usual, observing, that if the repetend of the number to be subtracted be greater than the repetend of the number it is to be taken from, then the right hand of the remainder must be less by unity than it would be if the expressions were finite. 1. Turn both the terms into their equivalent vulgar fractions, and find the product of those fractions as usual. 2. Turn the vulgar fraction expressing the product, into an equivalent decimal one, and it will be the product required. EXAMPLES. 1. Multiply 54 by 15. 544 and 15=11=3 084 the product. 2. Multiply 378-5 by 23.6. Ans. 8959.148. DIVISION OF CIRCULATING DECIMALS. RULE. 1. Change both the divisor and dividend into their equivalent vulgar fractions, and find their quotient as usual. 2. Turn the vulgar fraction expressing the quotient, into its equivalent decimal, and it will be the quotient required. A÷3=&×4=47=343=3-506493 the quotient. 2. Divide 345-8 by 6. Ans. 518.83. ALLIGATION IS the method of mixing two or more simples of different qualities, so that the composition may be of a mean or middle quality; It consists of two kinds, viz. Alligation Medial, and Alligation Al ternate. ALLIGATION MEDIAL Is, when the quantities and prices of several things are given, to find the mean price of the mixture compounded of those things. RULE. As the sum of the quantities, or the whole composition, is to their total value; so is any part of the composition to its mean price or value. EXAMPLES. 1. A Tobacconist would mix 60 of tobacco, at 6d. per with 5015 at 1s. 40 at 1s. 6d. and 30 at 2s. per : What is 1 of this mixture worth? £9. £ 60 at 0 6 is 1 10 As 180: 10:1 1 60 2. A farmer would mix 20 bushels of wheat at $1 per bushel, 16 bushels of rye at 75c. per bushel, 12 bushels of barley at 50c. per bushel, and 8 bushels of oats at 40c. per bushel: What is the value of one bushel of this mixture? Ans. 73c. 54m. 3. A wine merchant mixes 12 gallons of wine, at 75c. per gallon, with 24 gallons at 90c. and 16 gallons at $1 10c.: What is a gallon of this composition worth? Ans. 92c. 6m. 4. A goldsmith melted together 8oz. of gold of 22 carats fine. 1 8oz. of 21 carats fine, and 10oz. of 18 carats fine: Pray what is the quality, or fineness of the composition? 8x22+20×21+10×18, 8+20+10 =20, carats fine, Ans. 5. A refiner melts 55 of gold of 20 carats fine with 8 of 18 carats fine: How much alloy must be put to it, to make it 22 carats fine? 22-5×20+8×18÷÷÷5+8=3,3 Answer. It is not fine enough by 3 carats, so that no alloy must be added, but more gold. ALLIGATION ALTERNATE* Is the method of finding what quantity of each of the ingredients, whose rates are given, will compose a mixture of a given rate: So that it is the reverse of Alligation Medial, and may be proved by it. CASE I. The whole work of this case consists in linking the extremes truly together and taking the differences between them and the mean price, which differences are the quantities sought. RULE. 1. Place the several prices of the simples, being reduced to one denomination, in a column under each other, the least uppermost, and so gradually downward, as they increase with a line of *Demon. By connecting the less rate with the greater, and placing the dif ference between them and the mean rate alternately, or one after the other in turn, the quantities resulting are such, that there is precisely as much gained by one quantity as is lost by the other, and therefore the gain and loss, upon the whole, are equal, and are exactly the proposed rate. In like manner, let the number of simples be what it may, and with how many soever, each one is linked, since it is always a less with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequently an equal balance on the whole. It is obvious from the rule, that questions of this sort admit of a great variety of answers; for having found one answer, we may find as many more as we please, by only multiplying or dividing each of the quantities found, by 2, 3, 4, &c. the reason of which is evident; for if two quantities of two simples make a balance of loss and gain with respect to the mean price, so must also the double or triple, the half or third part, or any other ratio of these quantities, and so on ad infinitum. If any one of the simples be of little or no value with respect to the rest, it rate is supposed to be nothing, as water mixed with wine, and alloy with gold and silver. connection at the left hand, and the mean price at the left hand of all. 2. Connect, with a continued line, the price of each simple, or ingredient, which is less than that of the compound, with one or any number of those which are greater than the compound, and each greater rate or price with one or any number of the less. 3. Place the difference, between the mean price (or mixture rate) and that of each of the simples, opposite to the rates with which they are connected. 4. Then, if only one difference stand against any rate, it will be the quantity belonging to that rate; but if there be several, their sum will be the quantity. EXAMPLES. 1. A merchant has spices, some at 1s. 6d. per 1, some at 28. some at 4s. and some at 5s. per : How much of each sort must he mix that he may sell the mixture at 3s. 4d. per ib? per d. # s. d. s. d. 8 at 16 per it 16 Note. These seven answers arise from as many various ways of linking the rates of the ingredients together. 2. *A merchant has Canary wine, at 3s. per gallon, Sherry, at 25. 1d. and Claret at 1s. 5d. per gallon: How much of each sort must he take, to sell it at 2s. 4d. per gallon? 3+1114 at 3 0 36 Mean rate 28d. 25 17 8 8 21 per gallon. *Note, the 2d and 3d questions admit but of one way of linking, and so but of one answer; yet all numbers in the same proportion between themselves, as the numbers which compose the answer, will likewise satisfy the condition of the question. |